Psac manual

Annasaheb Dange College of Engineering & Technology, Ashta

Department of Electrical Engineering

Editor

Mr. Iranna M. Korachagaon

Assistant Professor – Electrical Engineering

Assistance

Mahesh P Gurav

February 2009, First Edition

Power System Analysis & Control

"SN "Title of the Experiment "Page No. "
"1 "Ybus Formation " "
"2 "Economic Load Dispatch " "
"3 "Symmetrical Components " "
"4 "Power flow solutions using Gauss Seidel Method " "
"5 "Power flow solutions using Newton Raphson " "
" "Method " "
"6 "Balance/ Unbalanced Fault Analysis " "
"7 "Equal Area Criterion " "
"8 "Transient Stability Analysis/ Swing Curve " "
"9 "Load Frequency Control " "
" "Instructions for Loading the MATLAB Programs " "

References:
[ AH] Electrical Power Systems by Asfaq Husain
[NK ] Power System Analysis by S.J. Nagrath and D.B. Kothari
[HS] Power System Analysis by Hadi Sadat
[BRG ] Power System Analysis & Control by B R Gupta


Assignment/ Experiment No. 1

"Subject: Power System Analysis & Control "
"Chapter: Ybus Formation "
"Name & Signature of the Faculty: Mr. Iranna Korachagaon "

A. Solve the following numerical

A.1.Find Ybus for the following bus system
[Ex.9.3, BRG]

"Line no.P-Q "1-2 "2-3 "3-4 "1-4 "1-3 "
"Rpu "0.025 "0.02 "0.05 "0.04 "0.1 "
"Xpu "0.1 "0.08 "0.2 "0.6 "0.4 "

A.2. Find Ybus for
[Ex. 14.6; AH]

"Bus code P-q "1-2 "1-3 "1-4 "2-3 "3-4 "
"Zpu "0.25+j1.0 "0.2+j0.8 "0.3+j1.2 "0.2+j0.8 "0.15+j0.6 "

A.3. Find Ybus for
[Ex. 9.4; BRG]

"Bus code P-Q "1-2 "2-3 "2-4 "3-4 "1-3 "
"Zpu "0.2+j0.8 "0.3+j0.9 "0.25+j1.0 "0.2+j0.8 "0.1+j0.4 "
"Charging "J0.02 "J0.03 "J0.04 "J0.02 "J0.01 "
"Admittance Ypq/2" " " " " "

A.4. Find Ybus for [Ex. 14.4; AH]

"Line Bus P-q "1-2 "1-3 "2-3 "
"Impedance Zpu "0.06+j0.18 "0.03+j0.09 "0.08+j0.24 "

B. Verify the computer solution for the following numerical.

B.1 The emf's shown in figure are E1=1.1 00 & E2=1 00. Find Y-bus & Z-
bus, & solve for the bus voltages.
[Ex. 6.1; HS]

Computer Program

>> % From To R X
z = [ 0 1 0 1.0
0 2 0 0.8
1 2 0 0.4
1 3 0 0.2
2 3 0 0.2
3 4 0 0.08];
[Ybus] = ybus(z) % bus admittance matrix
Ibus = [-j*1.1; -j*1.25; 0; 0]; % vector of injected bus currents
Zbus = inv(Ybus) % bus impedance matrix
Vbus = Zbus*Ibus
%Vbus = Ybus\Ibus
OUTPUT
Ybus =
0 - 8.5000i 0 + 2.5000i 0 + 5.0000i 0

0 + 2.5000i 0 - 8.7500i 0 + 5.0000i 0

0 + 5.0000i 0 + 5.0000i 0 -22.5000i 0
+12.5000i
0 0 0 +12.5000i 0
-12.5000i

Zbus =
0 + 0.5000i 0 + 0.4000i 0 + 0.4500i 0 +
0.4500i
0 + 0.4000i 0 + 0.4800i 0 + 0.4400i 0 +
0.4400i
0 + 0.4500i 0 + 0.4400i 0 + 0.5450i 0 +
0.5450i
0 + 0.4500i 0 + 0.4400i 0 + 0.5450i 0 +
0.6250i

Vbus =
1.0500
1.0400
1.0450
1.045
*** End of Experiment 1 ***


"Chapter: Economic Load Distribution "

C. Solve the following numerical

A.1. The fuel cost of two unit plant are given by
C1=100+2P1+0.05P12, C2=200+2P2+0.01P22, where P1 & P2 are in
MW. The plant supplies a load of 450 MW. Find the economic load
scheduling of the two units and the incremental fuel cost. Neglect
losses. [Ex. 15.1; AH]

A.2. Incremental fuel costs in Rs/MW for a plant operating 2 units are,
, Assume that both units are operating all the times &
total load varies from 40 MW to 250 MW. Maximum and minimum loads on
each unit are 125 MW & 20 MW respectively. How the load be the shared
between two units as system load varies over the full range.
[Ex. 7.1; NK]

A.3. With the data in Ex.2, find the saving in fuel cost in Rs/hr. For
optimum scheduling of the
total load of 130 MW.
[Ex. 7.2; NK]

A.4. A two bus system is shown. If 100 MW is transmitted from plant 1, to
the load, a transmission loss of 10 MW is ineurred. Find the required
generation for each plant & power received by load when the system
lamda(λ) is Rs.25/MWh. The incremental fuel cost of the two plants are
given below,
; Rs/MWH; [Ex.7.4;NK]

A.5. Consider the system in Ex. 4, with a load of 237.04 MW at bus 2,find
the optimum load distribution between the two plats for i) When losses
are included but not co-ordinated ii) When losses are also co-
ordinated. Also find the saving in Rs/hr. when losses are co-
ordinated. Generator limits find the optimal dispatch and the total
cost in Rs./h
[Ex.
7.5;NK]


B.1 The fuel cost functions for 3 thermal plants in Rs./h are given
by,
C1=500+5.3P1+0.004P12, C2=400+5.5P2+0.006P22,
C3=200+5.8P3+0.009P32
where P1, P2 and P3 are in MW. The total load Pd is 800MW. Neglecting
line losses and generator limits find the optimal dispatch and the
total cost in Rs./h. (Answer: P1=400, P2=250, P3=150, lambda=8.5,
Total Cost per hour = Rs. 6682.5)
[Ex. 7.4; HS]
Computer Program

% Iterative solution Using Newton method
alpha =[500; 400; 200];
beta = [5.3; 5.5; 5.8]; gama=[.004; .006; .009];
PD=800;
DelP = 10; % Error in DelP is set to a high value
lambda = input('Enter estimated value of Lambda = ');
fprintf('\n ')
disp([' Lambda P1 P2 P3 DP'...
' grad Delambda'])
iter = 0; % Iteration counter
while abs(DelP) >= 0.001 % Test for convergence
iter = iter + 1; % No. of iterations
P = (lambda - beta)./(2*gama);
DelP =PD - sum(P); % Residual
J = sum( ones(length(gama), 1)./(2*gama)); % Gradient sum
Delambda = DelP/J; % Change in variable
disp([lambda, P(1), P(2), P(3), DelP, J, Delambda])
lambda = lambda + Delambda; % Successive solution
end
totalcost = sum(alpha + beta.*P + gama.*P.^2)
%Graphical Demonstration of Example 7.4
axis([0 450 6.5 10.5]);
P1=250:10:450; P2 = 150:10:350; P3=100:10:250;
IC1= 5.3 + 0.008*P1;
IC2= 5.5 + 0.012*P2;
IC3= 5.8 + 0.018*P3;
Px = 0:100:400;
plot(P1, IC1, P2, IC2, P3, IC3, Px, lambda*ones(1, length(Px)),'-m'),
xlabel('P, MW'), ylabel(' $/MWh'), grid

Output
Enter estimated value of Lambda = 8.5
Lambda P1 P2 P3 DP
grad Delambda
8.5000 400.0000 250.0000 150.0000 0
263.8889 0
totalcost =
6.6825e+003

Analyze the output graph:
B.2 The fuel cost functions for 3 thermal plants inRs./h are given by,
C1=200+7.0P1+0.008P12, C2=180+6.3P2+0.009P22,
C3=140+6.8P3+0.007P32
Where P1, P2 and P3 are in MW. Plant outputs are 10 MW< 85 MW, 10MW< 80 MW,
10MW< 70 MW The total load Pd is 150 MW. Determine the optimal dispatch of
generation and the total cost in Rs./h. (Answer: P1=35.0907, P2=64.1317,
P3=52.4767, Total genaration Cost per hour = Rs. 1592.65). [Ex. 7.7; HS]

% This program solves the coordination equation for economic scheduling
% of generation. The program requires the total load demand (Pdt), the
% cost function matrix (cost) and the gen. mwlimits. If mwlimits is not
% defined the scheduling is obtained with no generation limits. If the
% basemva and any of the loss coefficients B, B0 and B00 are specified
% optimum dispatch is obtained including the system losses.
%
% copyright (c) 1998 by H. Saadat

clear Pgg
if exist('Pdt')~=1
Pdt = input('Enter total demand Pdt = ');
else, end
if exist('cost')~=1
cost = input('Enter the cost matrix, cost = ');
else, end
ngg = length(cost(:,1));
if exist('mwlimits')~=1
mwlimits= [zeros(ngg, 1), inf*ones(ngg,1)];
else, end
if exist('B')~=1
B = zeros(ngg, ngg);
else, end
if exist('B0')~=1
B0=zeros(1, ngg);
else, end
if exist('B00')~=1
B00=0;
else, end
if exist('basemva')~=1
basemva=100;
else, end
clear Pgg
Bu=B/basemva; B00u=basemva*B00;
alpha=cost(:,1); beta=cost(:,2); gama = cost(:,3);
Pmin=mwlimits(:,1); Pmax=mwlimits(:,2);
wgt=ones(1, ngg);
if Pdt > sum(Pmax)
Error1 = ['Total demand is greater than the total sum of maximum
generation.'
'No feasible solution. Reduce demand or correct generator
limits.'];
disp(Error1), return
elseif Pdt < sum(Pmin)
Error2 = ['Total demand is less than the total sum of minimum generation.
'
'No feasible solution. Increase demand or correct generator
limits.'];
disp(Error2), return
else, end
iterp = 0; % Iteration counter
DelP = 10; % Error in DelP is set to a high value

E=Bu;
if exist('lambda')~=1
lambda=max(beta);
end
while abs(DelP) >= 0.0001 & iterp < 200 % Test for convergence
iterp = iterp + 1; % No. of iterations
for k=1:ngg
if wgt(k) == 1
E(k,k) = gama(k)/lambda + Bu(k,k);
Dx(k) = 1/2*(1 - B0(k)- beta(k)/lambda);
else, E(k,k)=1; Dx(k) = 0;
for m=1:ngg
if m~=k
E(k,m)=0;
else,end
end
end
end
PP=E\Dx';
for k=1:ngg
if wgt(k)==1
Pgg(k) = PP(k);
else,end
end
Pgtt = sum(Pgg);
PL=Pgg*Bu*Pgg'+B0*Pgg'+B00u;
DelP =Pdt+PL -Pgtt ; %Residual
for k = 1:ngg
if Pgg(k) > Pmax(k) & abs(DelP) <=0.001,
Pgg(k) = Pmax(k); wgt(k) = 0;
elseif Pgg(k) < Pmin(k) & abs(DelP) <= 0.001
Pgg(k) = Pmin(k); wgt(k) = 0;
else, end
end
PL=Pgg*Bu*Pgg'+B0*Pgg'+B00u;
DelP =Pdt +PL - sum(Pgg); %Residual
for k=1:ngg
BP = 0;
for m=1:ngg
if m~=k
BP = BP + Bu(k,m)*Pgg(m);
else, end
end
grad(k)=(gama(k)*(1-B0(k))+Bu(k,k)*beta(k)-
2*gama(k)*BP)/(2*(gama(k)+lambda*Bu(k,k))^2);
end
sumgrad=wgt*grad';
Delambda = DelP/sumgrad; % Change in variable
lambda = lambda + Delambda; % Successive solution
end
fprintf('Incremental cost of delivered power (system lambda) = %9.6f
Rs/MWh \n', lambda);
fprintf('Optimal Dispatch of Generation:\n\n')
disp(Pgg')
%fprintf('Total system loss = %g MW \n\n', PL)
ng=length(Pgg);
n=0;
if exist('nbus')==1 " exist('busdata')==1
for k=1:nbus
if kb(k)~=0
n=n+1;
if n <= ng
busdata(k,7)=Pgg(n); else, end
else , end
end
if n == ng
for k=1:nbus
if kb(k)==1
dpslack = abs(Pg(k)-busdata(k,7))/basemva;
fprintf('Absolute value of the slack bus real power mismatch,
dpslack = %8.4f pu \n', dpslack)
else, end
end
else, end
else, end
clear BP Dx DelP Delambda E PP grad sumgrad wgt Bu B00u B B0 B00

% This program computes the total generation cost. It requires the
% real power generation schedule and the cost matrix.
% Copyright (C) 1998 by H. Saadat

%function [totalcost]=gencost(Pgg, cost)
if exist('Pgg')~=1
Pgg=input('Enter the scheduled real power gen. in row matrix ');
else,end
if exist('cost')~=1
cost = input('Enter the cost function matrix ');
else, end
ngg = length(cost(:,1));
Pmt = [ones(1,ngg); Pgg; Pgg.^2];
for i = 1:ngg
costv(i) = cost(i,:)*Pmt(:,i);
end
totalcost=sum(costv);
fprintf('\nTotal generation cost = % 10.2f Rs. /h \n', totalcost)
cost = [200 7.0 0.008
180 6.3 0.009
140 6.8 0.007];
mwlimits =[10 85
10 80
10 70];
Pdt = 150;
B = [0.0218 0 0
0 0.0228 0
0 0 0.0179];
basemva = 100;
dispatch
gencost

Output
Incremental cost of delivered power (system lambda) = 7.510995 Rs. /MWh
Optimal Dispatch of Generation:

31.9372
67.2775
50.7853

Total generation cost = 1579.70 Rs./h




"Chapter: Symmetrical Components "


A.1. Determine the symmetrical components of currents, in three phase
system. The original phasors are Ia=12+j6, Ib=12-j2, Ic=-
15+j10. [Ex. 17.2;AH]

A.2. The line currents in a 3-phase system are Ia=5 600, Ib=5 -600 and
Ic=0. Find the symmetrical components.
[Ex. 10.6;BRG]


B.1 Obtain the symmetrical components of a set of unbalanced currents
Ia=1.6 250, Ib=100 1800, Ic=0.9 1320

(Ans:-Io=0.45 96.45290, I1=0.9435 -0.05500, I2=0.6024
22.31570)

[Ex. 10.1; HS]

Computer Program

Iabc = [1.6 25
1.0 180
.9 132];
I012 = abc2sc(Iabc); % Symmetrical components of phase a
I012p= rec2pol(I012) % Converts rectangular phasors into polar form

Output
I012p =

0.4512 96.4529
0.9435 -0.0550
0.6024 22.3157

B.2 The symmetrical components of a set of unbalanced 3-phase voltages for
phase 'a' are Va0=0.6 900, Va1=1.0 300, Va2=0.8 -300 . obtain the original
phasors.

(Ans.:Va=1.7088 24.18250, Vb=0.40 900, Vc=1.7088 155081750)

[Ex. 10.2;HS]

Computer Program

V012 = [.6 90
1 30
0.8 -30];
Vabc = sc2abc(V012); % Unbalanced phasors from symmetrical components
Vabcp= rec2pol(Vabc) % Converts rectangular phasors into polar form

Output
Vabcp =

1.7088 24.1825
0.4000 90.0000
1.7088 155.8175




"Chapter: Power flow Solution by G-S Method "


A.1. A three phase bus system is shown in below. The relevant pu line
admittance on 100 MVA base are indicated. Bus data are given in the
table. Form y-bus & determine the voltage at bus 2 & 3 after the second
iteration using Gauss-Siedel method. Take acceleration factor α=1.6.
Solve for only one iteration.

"Bus No. "type "Generato" "load " "Bus " "
" " "r " " " "voltage " "
" " "PG(MW) "QG(MVAR) "PL(MW) "QL(MVAR) "Vpu "δ "
"1 "slack "? "? "0 "0 "1.02 "0 "
"2 "PQ "25 "15 "50 "25 "? "?? "
"3 "PQ "0 "0 "60 "30 "? "? "

[ Ex. 14.8; AH]


B.1 30 bus IEEE sample system computer program is given analyze the
output.

[Ex. 6.9, HS]

Computer Program

clear
basemva = 100; accuracy = 0.001; accel = 1.8; maxiter = 100;

% IEEE 30-BUS TEST SYSTEM (American Electric Power)
% Bus Bus Voltage Angle ---Load---- -------Generator----- Static
Mvar
% No code Mag. Degree MW Mvar MW Mvar Qmin Qmax +Qc/-Ql
busdata=[1 1 1.06 0.0 0.0 0.0 0.0 0.0 0 0 0
2 2 1.043 0.0 21.70 12.7 40.0 0.0 -40 50 0
3 0 1.0 0.0 2.4 1.2 0.0 0.0 0 0 0
4 0 1.06 0.0 7.6 1.6 0.0 0.0 0 0 0
5 2 1.01 0.0 94.2 19.0 0.0 0.0 -40 40 0
6 0 1.0 0.0 0.0 0.0 0.0 0.0 0 0 0
7 0 1.0 0.0 22.8 10.9 0.0 0.0 0 0 0
8 2 1.01 0.0 30.0 30.0 0.0 0.0 -30 40 0
9 0 1.0 0.0 0.0 0.0 0.0 0.0 0 0 0
10 0 1.0 0.0 5.8 2.0 0.0 0.0 -6 24 19
11 2 1.082 0.0 0.0 0.0 0.0 0.0 0 0 0
12 0 1.0 0 11.2 7.5 0 0 0 0 0
13 2 1.071 0 0 0.0 0 0 -6 24 0
14 0 1 0 6.2 1.6 0 0 0 0 0
15 0 1 0 8.2 2.5 0 0 0 0 0
16 0 1 0 3.5 1.8 0 0 0 0 0
17 0 1 0 9.0 5.8 0 0 0 0 0
18 0 1 0 3.2 0.9 0 0 0 0 0
19 0 1 0 9.5 3.4 0 0 0 0 0
20 0 1 0 2.2 0.7 0 0 0 0 0
21 0 1 0 17.5 11.2 0 0 0 0 0
22 0 1 0 0 0.0 0 0 0 0 0
23 0 1 0 3.2 1.6 0 0 0 0 0
24 0 1 0 8.7 6.7 0 0 0 0 4.3
25 0 1 0 0 0.0 0 0 0 0 0
26 0 1 0 3.5 2.3 0 0 0 0 0
27 0 1 0 0 0.0 0 0 0 0 0
28 0 1 0 0 0.0 0 0 0 0 0
29 0 1 0 2.4 0.9 0 0 0 0 0
30 0 1 0 10.6 1.9 0 0 0 0 0];

% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr p.u. p.u. p.u. > 1 or < 1 tr. tap at bus nl
linedata=[1 2 0.0192 0.0575 0.02640 1
1 3 0.0452 0.1852 0.02040 1
2 4 0.0570 0.1737 0.01840 1
3 4 0.0132 0.0379 0.00420 1
2 5 0.0472 0.1983 0.02090 1
2 6 0.0581 0.1763 0.01870 1
4 6 0.0119 0.0414 0.00450 1
5 7 0.0460 0.1160 0.01020 1
6 7 0.0267 0.0820 0.00850 1
6 8 0.0120 0.0420 0.00450 1
6 9 0.0 0.2080 0.0 0.978
6 10 0 .5560 0 0.969
9 11 0 .2080 0 1
9 10 0 .1100 0 1
4 12 0 .2560 0 0.932
12 13 0 .1400 0 1
12 14 .1231 .2559 0 1
12 15 .0662 .1304 0 1
12 16 .0945 .1987 0 1
14 15 .2210 .1997 0 1
16 17 .0824 .1923 0 1
15 18 .1073 .2185 0 1
18 19 .0639 .1292 0 1
19 20 .0340 .0680 0 1
10 20 .0936 .2090 0 1
10 17 .0324 .0845 0 1
10 21 .0348 .0749 0 1
10 22 .0727 .1499 0 1
21 22 .0116 .0236 0 1
15 23 .1000 .2020 0 1
22 24 .1150 .1790 0 1
23 24 .1320 .2700 0 1
24 25 .1885 .3292 0 1
25 26 .2544 .3800 0 1
25 27 .1093 .2087 0 1
28 27 0 .3960 0 0.968
27 29 .2198 .4153 0 1
27 30 .3202 .6027 0 1
29 30 .2399 .4533 0 1
8 28 .0636 .2000 0.0214 1
6 28 .0169 .0599 0.065 1];

lfybus % form the bus admittance matrix
lfgauss % Load flow solution by Gauss-Seidel method
busout % Prints the power flow solution on the screen
lineflow % Computes and displays the line flow and losses
Power Flow Solution by Gauss-Seidel Method
Maximum Power Mismatch = 0.000951884
No. of Iterations = 34

Bus Voltage Angle ------Load------ ---Generation--- Injected
No. Mag. Degree MW Mvar MW Mvar Mvar

1 1.060 0.000 0.000 0.000 260.950 -17.010 0.000
2 1.043 -5.496 21.700 12.700 40.000 48.826 0.000
3 1.022 -8.002 2.400 1.200 0.000 0.000 0.000
4 1.013 -9.659 7.600 1.600 0.000 0.000 0.000
5 1.010 -14.380 94.200 19.000 0.000 35.995 0.000
6 1.012 -11.396 0.000 0.000 0.000 0.000 0.000
7 1.003 -13.149 22.800 10.900 0.000 0.000 0.000
8 1.010 -12.114 30.000 30.000 0.000 30.759 0.000
9 1.051 -14.432 0.000 0.000 0.000 0.000 0.000
10 1.044 -16.024 5.800 2.000 0.000 0.000 19.000
11 1.082 -14.432 0.000 0.000 0.000 16.113 0.000
12 1.057 -15.301 11.200 7.500 0.000 0.000 0.000
13 1.071 -15.300 0.000 0.000 0.000 10.406 0.000
14 1.043 -16.190 6.200 1.600 0.000 0.000 0.000
15 1.038 -16.276 8.200 2.500 0.000 0.000 0.000
16 1.045 -15.879 3.500 1.800 0.000 0.000 0.000
17 1.039 -16.187 9.000 5.800 0.000 0.000 0.000
18 1.028 -16.881 3.200 0.900 0.000 0.000 0.000
19 1.025 -17.049 9.500 3.400 0.000 0.000 0.000
20 1.029 -16.851 2.200 0.700 0.000 0.000 0.000
21 1.032 -16.468 17.500 11.200 0.000 0.000 0.000
22 1.033 -16.455 0.000 0.000 0.000 0.000 0.000
23 1.027 -16.660 3.200 1.600 0.000 0.000 0.000
24 1.022 -16.829 8.700 6.700 0.000 0.000 4.300
25 1.019 -16.423 0.000 0.000 0.000 0.000 0.000
26 1.001 -16.835 3.500 2.300 0.000 0.000 0.000
27 1.026 -15.913 0.000 0.000 0.000 0.000 0.000
28 1.011 -12.056 0.000 0.000 0.000 0.000 0.000
29 1.006 -17.133 2.400 0.900 0.000 0.000 0.000
30 0.994 -18.016 10.600 1.900 0.000 0.000 0.000

Total 283.400 126.200 300.950 125.089 23.300

Line Flow and Losses

--Line-- Power at bus & line flow --Line loss-- Transformer
from to MW Mvar MVA MW Mvar tap

1 260.950 -17.010 261.504
2 177.743 -22.140 179.117 5.461 10.517
3 83.197 5.125 83.354 2.807 7.079

2 18.300 36.126 40.497
1 -172.282 32.657 175.350 5.461 10.517
4 45.702 2.720 45.783 1.106 -0.519
5 82.990 1.704 83.008 2.995 8.178
6 61.905 -0.966 61.913 2.047 2.263

3 -2.400 -1.200 2.683
1 -80.390 1.954 80.414 2.807 7.079
4 78.034 -3.087 78.095 0.771 1.345

4 -7.600 -1.600 7.767
2 -44.596 -3.239 44.713 1.106 -0.519
3 -77.263 4.432 77.390 0.771 1.345
6 70.132 -17.624 72.313 0.605 1.181
12 44.131 14.627 46.492 0.000 4.686 0.932

5 -94.200 16.995 95.721
2 -79.995 6.474 80.256 2.995 8.178
7 -14.210 10.467 17.649 0.151 -1.687

6 0.000 0.000 0.000
2 -59.858 3.229 59.945 2.047 2.263
4 -69.527 18.805 72.026 0.605 1.181
7 37.537 -1.915 37.586 0.368 -0.598
8 29.534 -3.712 29.766 0.103 -0.558
9 27.687 -7.318 28.638 0.000 1.593 0.978
10 15.828 0.656 15.842 -0.000 1.279 0.969
28 18.840 -9.575 21.134 0.060 -13.085

7 -22.800 -10.900 25.272
5 14.361 -12.154 18.814 0.151 -1.687
6 -37.170 1.317 37.193 0.368 -0.598

8 -30.000 0.759 30.010
6 -29.431 3.154 29.599 0.103 -0.558
28 -0.570 -2.366 2.433 0.000 -4.368

9 0.000 0.000 0.000
6 -27.687 8.911 29.086 0.000 1.593
11 0.003 -15.653 15.653 0.000 0.461
10 27.731 6.747 28.540 -0.000 0.811

10 -5.800 17.000 17.962
6 -15.828 0.623 15.840 -0.000 1.279
9 -27.731 -5.936 28.359 -0.000 0.811
20 9.018 3.569 9.698 0.081 0.180
17 5.347 4.393 6.920 0.014 0.037
21 15.723 9.846 18.551 0.110 0.236
22 7.582 4.487 8.811 0.052 0.107

11 0.000 16.113 16.113
9 -0.003 16.114 16.114 0.000 0.461

12 -11.200 -7.500 13.479
4 -44.131 -9.941 45.237 0.000 4.686
13 -0.021 -10.274 10.274 0.000 0.132
14 7.852 2.428 8.219 0.074 0.155
15 17.852 6.968 19.164 0.217 0.428
16 7.206 3.370 7.955 0.053 0.112

13 0.000 10.406 10.406
12 0.021 10.406 10.406 0.000 0.132

14 -6.200 -1.600 6.403
12 -7.778 -2.273 8.103 0.074 0.155
15 1.592 0.708 1.742 0.006 0.006

15 -8.200 -2.500 8.573
12 -17.634 -6.540 18.808 0.217 0.428
14 -1.586 -0.702 1.734 0.006 0.006
18 6.009 1.741 6.256 0.039 0.079
23 5.004 2.963 5.815 0.031 0.063

16 -3.500 -1.800 3.936
12 -7.152 -3.257 7.859 0.053 0.112
17 3.658 1.440 3.931 0.012 0.027

17 -9.000 -5.800 10.707
16 -3.646 -1.413 3.910 0.012 0.027
10 -5.332 -4.355 6.885 0.014 0.037

18 -3.200 -0.900 3.324
15 -5.970 -1.661 6.197 0.039 0.079
19 2.779 0.787 2.888 0.005 0.010

19 -9.500 -3.400 10.090
18 -2.774 -0.777 2.881 0.005 0.010
20 -6.703 -2.675 7.217 0.017 0.034

20 -2.200 -0.700 2.309
19 6.720 2.709 7.245 0.017 0.034
10 -8.937 -3.389 9.558 0.081 0.180

21 -17.500 -11.200 20.777
10 -15.613 -9.609 18.333 0.110 0.236
22 -1.849 -1.627 2.463 0.001 0.001

22 0.000 0.000 0.000
10 -7.531 -4.380 8.712 0.052 0.107
21 1.850 1.628 2.464 0.001 0.001
24 5.643 2.795 6.297 0.043 0.067

23 -3.200 -1.600 3.578
15 -4.972 -2.900 5.756 0.031 0.063
24 1.771 1.282 2.186 0.006 0.012

24 -8.700 -2.400 9.025
22 -5.601 -2.728 6.230 0.043 0.067
23 -1.765 -1.270 2.174 0.006 0.012
25 -1.322 1.604 2.079 0.008 0.014

25 0.000 0.000 0.000
24 1.330 -1.590 2.073 0.008 0.014
26 3.520 2.372 4.244 0.044 0.066
27 -4.866 -0.786 4.929 0.026 0.049

26 -3.500 -2.300 4.188
25 -3.476 -2.306 4.171 0.044 0.066

27 0.000 0.000 0.000
25 4.892 0.835 4.963 0.026 0.049
28 -18.192 -4.152 18.660 -0.000 1.310
29 6.178 1.675 6.401 0.086 0.162
30 7.093 1.663 7.286 0.162 0.304

28 0.000 0.000 0.000
27 18.192 5.463 18.994 -0.000 1.310 0.968
8 0.570 -2.003 2.082 0.000 -4.368
6 -18.780 -3.510 19.106 0.060 -13.085

29 -2.400 -0.900 2.563
27 -6.093 -1.513 6.278 0.086 0.162
30 3.716 0.601 3.764 0.034 0.063

30 -10.600 -1.900 10.769
27 -6.932 -1.359 7.064 0.162 0.304
29 -3.683 -0.537 3.722 0.034 0.063

Total loss 17.594 22.233




"Chapter: Power Flow Solution by N-R Method "


A.1. Figure below shows a 3-bus system. The series impedance & shunt
admittance of each line are 0.026+ j 0.11 pu & j0.04 pu respectively.
The bus data is given in table

"BUS "PG "QG "PL "QL "Bus " "
" " " " " "voltage " "
"1 "? "? "1.0 "0.5 "1.03+j0 "Slack bus "
"2 "1.5 "? "0 "0 "V=1.03 "PV bus "
"3 "0 "0 "1.2 "0.5 "? "PQ bus "

For bus 2 the minimum & maximum reactance power limits are 0 & 0.8 pu. (a)
form [Y-bus]

(b) Find P20, Q2o, P30, Q30 (c) Find[ J] (d)Form the
general equation for calculating the changes in variables by N-R method.
Write the equations for the first iteration.

[ Ex. 9.16, BRG]


B.1. Analysis the power flow solution obtained by Newton-Rapson method. A
computer program is given for 30-bus IEEE test system.

[Ex. 6.11;HS]

Computer Program

>> clear
basemva = 100; accuracy = 0.001; maxiter = 10;

% IEEE 30-BUS TEST SYSTEM (American Electric Power)
% Bus Bus Voltage Angle ---Load---- -------Generator-----
Injected
% No code Mag. Degree MW Mvar MW Mvar Qmin Qmax Mvar
busdata=[1 1 1.06 0.0 0.0 0.0 0.0 0.0 0 0 0
2 2 1.043 0.0 21.70 12.7 40.0 0.0 -40 50 0
3 0 1.0 0.0 2.4 1.2 0.0 0.0 0 0 0
4 0 1.06 0.0 7.6 1.6 0.0 0.0 0 0 0
5 2 1.01 0.0 94.2 19.0 0.0 0.0 -40 40 0
6 0 1.0 0.0 0.0 0.0 0.0 0.0 0 0 0
7 0 1.0 0.0 22.8 10.9 0.0 0.0 0 0 0
8 2 1.01 0.0 30.0 30.0 0.0 0.0 -10 60 0
9 0 1.0 0.0 0.0 0.0 0.0 0.0 0 0 0
10 0 1.0 0.0 5.8 2.0 0.0 0.0 -6 24 19
11 2 1.082 0.0 0.0 0.0 0.0 0.0 0 0 0
12 0 1.0 0 11.2 7.5 0 0 0 0 0
13 2 1.071 0 0 0.0 0 0 -6 24 0
14 0 1 0 6.2 1.6 0 0 0 0 0
15 0 1 0 8.2 2.5 0 0 0 0 0
16 0 1 0 3.5 1.8 0 0 0 0 0
17 0 1 0 9.0 5.8 0 0 0 0 0
18 0 1 0 3.2 0.9 0 0 0 0 0
19 0 1 0 9.5 3.4 0 0 0 0 0
20 0 1 0 2.2 0.7 0 0 0 0 0
21 0 1 0 17.5 11.2 0 0 0 0 0
22 0 1 0 0 0.0 0 0 0 0 0
23 0 1 0 3.2 1.6 0 0 0 0 0
24 0 1 0 8.7 6.7 0 0 0 0 4.3
25 0 1 0 0 0.0 0 0 0 0 0
26 0 1 0 3.5 2.3 0 0 0 0 0
27 0 1 0 0 0.0 0 0 0 0 0
28 0 1 0 0 0.0 0 0 0 0 0
29 0 1 0 2.4 0.9 0 0 0 0 0
30 0 1 0 10.6 1.9 0 0 0 0 0];

% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr p.u. p.u. p.u. > 1 or < 1 tr. tap at bus nl
linedata=[1 2 0.0192 0.0575 0.02640 1
1 3 0.0452 0.1852 0.02040 1
2 4 0.0570 0.1737 0.01840 1
3 4 0.0132 0.0379 0.00420 1
2 5 0.0472 0.1983 0.02090 1
2 6 0.0581 0.1763 0.01870 1
4 6 0.0119 0.0414 0.00450 1
5 7 0.0460 0.1160 0.01020 1
6 7 0.0267 0.0820 0.00850 1
6 8 0.0120 0.0420 0.00450 1
6 9 0.0 0.2080 0.0 0.978
6 10 0 .5560 0 0.969
9 11 0 .2080 0 1
9 10 0 .1100 0 1
4 12 0 .2560 0 0.932
12 13 0 .1400 0 1
12 14 .1231 .2559 0 1
12 15 .0662 .1304 0 1
12 16 .0945 .1987 0 1
14 15 .2210 .1997 0 1
16 17 .0824 .1923 0 1
15 18 .1073 .2185 0 1
18 19 .0639 .1292 0 1
19 20 .0340 .0680 0 1
10 20 .0936 .2090 0 1
10 17 .0324 .0845 0 1
10 21 .0348 .0749 0 1
10 22 .0727 .1499 0 1
21 22 .0116 .0236 0 1
15 23 .1000 .2020 0 1
22 24 .1150 .1790 0 1
23 24 .1320 .2700 0 1
24 25 .1885 .3292 0 1
25 26 .2544 .3800 0 1
25 27 .1093 .2087 0 1
28 27 0 .3960 0 0.968
27 29 .2198 .4153 0 1
27 30 .3202 .6027 0 1
29 30 .2399 .4533 0 1
8 28 .0636 .2000 0.0214 1
6 28 .0169 .0599 0.065 1];

lfybus % form the bus admittance matrix
lfnewton % Load flow solution by Newton-Raphson method
busout % Prints the power flow solution on the screen
lineflow % computes and displays the line flow and losses
Power Flow Solution by Newton-Raphson Method
Maximum Power Mismatch = 7.54898e-007
No. of Iterations = 4

Bus Voltage Angle ------Load------ ---Generation--- Injected
No. Mag. Degree MW Mvar MW Mvar Mvar

1 1.060 0.000 0.000 0.000 260.998 -17.021 0.000
2 1.043 -5.497 21.700 12.700 40.000 48.822 0.000
3 1.022 -8.004 2.400 1.200 0.000 0.000 0.000
4 1.013 -9.661 7.600 1.600 0.000 0.000 0.000
5 1.010 -14.381 94.200 19.000 0.000 35.975 0.000
6 1.012 -11.398 0.000 0.000 0.000 0.000 0.000
7 1.003 -13.150 22.800 10.900 0.000 0.000 0.000
8 1.010 -12.115 30.000 30.000 0.000 30.826 0.000
9 1.051 -14.434 0.000 0.000 0.000 0.000 0.000
10 1.044 -16.024 5.800 2.000 0.000 0.000 19.000
11 1.082 -14.434 0.000 0.000 0.000 16.119 0.000
12 1.057 -15.302 11.200 7.500 0.000 0.000 0.000
13 1.071 -15.302 0.000 0.000 0.000 10.423 0.000
14 1.042 -16.191 6.200 1.600 0.000 0.000 0.000
15 1.038 -16.278 8.200 2.500 0.000 0.000 0.000
16 1.045 -15.880 3.500 1.800 0.000 0.000 0.000
17 1.039 -16.188 9.000 5.800 0.000 0.000 0.000
18 1.028 -16.884 3.200 0.900 0.000 0.000 0.000
19 1.025 -17.052 9.500 3.400 0.000 0.000 0.000
20 1.029 -16.852 2.200 0.700 0.000 0.000 0.000
21 1.032 -16.468 17.500 11.200 0.000 0.000 0.000
22 1.033 -16.455 0.000 0.000 0.000 0.000 0.000
23 1.027 -16.662 3.200 1.600 0.000 0.000 0.000
24 1.022 -16.830 8.700 6.700 0.000 0.000 4.300
25 1.019 -16.424 0.000 0.000 0.000 0.000 0.000
26 1.001 -16.842 3.500 2.300 0.000 0.000 0.000
27 1.026 -15.912 0.000 0.000 0.000 0.000 0.000
28 1.011 -12.057 0.000 0.000 0.000 0.000 0.000
29 1.006 -17.136 2.400 0.900 0.000 0.000 0.000
30 0.995 -18.015 10.600 1.900 0.000 0.000 0.000

Total 283.400 126.200 300.998 125.144 23.300

Line Flow and Losses
--Line-- Power at bus & line flow --Line loss-- Transformer
from to MW Mvar MVA MW Mvar tap

1 260.998 -17.021 261.553
2 177.778 -22.148 179.152 5.464 10.524
3 83.221 5.127 83.378 2.808 7.085

2 18.300 36.122 40.493
1 -172.314 32.671 175.384 5.464 10.524
4 45.712 2.705 45.792 1.106 -0.517
5 82.990 1.703 83.008 2.995 8.178
6 61.912 -0.958 61.920 2.048 2.264

3 -2.400 -1.200 2.683
1 -80.412 1.958 80.436 2.808 7.085
4 78.012 -3.158 78.076 0.771 1.344

4 -7.600 -1.600 7.767
2 -44.605 -3.222 44.722 1.106 -0.517
3 -77.242 4.503 77.373 0.771 1.344
6 70.126 -17.526 72.282 0.604 1.179
12 44.121 14.646 46.489 0.000 4.685 0.932

5 -94.200 16.975 95.717
2 -79.995 6.475 80.257 2.995 8.178
7 -14.205 10.500 17.664 0.151 -1.687

6 0.000 0.000 0.000
2 -59.864 3.222 59.951 2.048 2.264
4 -69.521 18.705 71.994 0.604 1.179
7 37.523 -1.885 37.570 0.367 -0.598
8 29.528 -3.754 29.766 0.103 -0.558
9 27.693 -7.322 28.644 -0.000 1.594 0.978
10 15.823 0.653 15.836 0.000 1.278 0.969
28 18.819 -9.618 21.134 0.060 -13.086

7 -22.800 -10.900 25.272
5 14.356 -12.187 18.831 0.151 -1.687
6 -37.156 1.287 37.178 0.367 -0.598

8 -30.000 0.826 30.011
6 -29.425 3.196 29.598 0.103 -0.558
28 -0.575 -2.370 2.438 0.000 -4.368

9 0.000 0.000 0.000
6 -27.693 8.916 29.093 -0.000 1.594
11 -0.000 -15.657 15.657 0.000 0.462
10 27.693 6.741 28.501 0.000 0.809

10 -5.800 17.000 17.962
6 -15.823 0.626 15.835 0.000 1.278
9 -27.693 -5.932 28.321 0.000 0.809
20 9.027 3.560 9.704 0.081 0.180
17 5.372 4.414 6.953 0.014 0.037
21 15.733 9.842 18.558 0.110 0.237
22 7.583 4.490 8.813 0.052 0.107

11 0.000 16.119 16.119
9 0.000 16.119 16.119 0.000 0.462

12 -11.200 -7.500 13.479
4 -44.121 -9.961 45.232 0.000 4.685
13 0.000 -10.291 10.291 -0.000 0.133
14 7.856 2.442 8.227 0.075 0.155
15 17.857 6.947 19.161 0.217 0.428
16 7.208 3.363 7.954 0.053 0.112

13 0.000 10.423 10.423
12 -0.000 10.424 10.424 -0.000 0.133

14 -6.200 -1.600 6.403
12 -7.782 -2.287 8.111 0.075 0.155
15 1.582 0.687 1.724 0.006 0.005

15 -8.200 -2.500 8.573
12 -17.640 -6.519 18.806 0.217 0.428
14 -1.576 -0.681 1.717 0.006 0.005
18 6.014 1.744 6.262 0.039 0.080
23 5.001 2.956 5.810 0.031 0.063

16 -3.500 -1.800 3.936
12 -7.154 -3.251 7.858 0.053 0.112
17 3.654 1.451 3.932 0.012 0.027

17 -9.000 -5.800 10.707
16 -3.643 -1.424 3.911 0.012 0.027
10 -5.357 -4.376 6.918 0.014 0.037

18 -3.200 -0.900 3.324
15 -5.975 -1.665 6.203 0.039 0.080
19 2.775 0.765 2.879 0.005 0.010

19 -9.500 -3.400 10.090
18 -2.770 -0.755 2.871 0.005 0.010
20 -6.730 -2.645 7.231 0.017 0.034

20 -2.200 -0.700 2.309
19 6.747 2.679 7.259 0.017 0.034
10 -8.947 -3.379 9.564 0.081 0.180

21 -17.500 -11.200 20.777
10 -15.623 -9.606 18.340 0.110 0.237
22 -1.877 -1.594 2.462 0.001 0.001

22 0.000 0.000 0.000
10 -7.531 -4.384 8.714 0.052 0.107
21 1.877 1.596 2.464 0.001 0.001
24 5.654 2.788 6.304 0.043 0.067

23 -3.200 -1.600 3.578
15 -4.970 -2.893 5.751 0.031 0.063
24 1.770 1.293 2.192 0.006 0.012

24 -8.700 -2.400 9.025
22 -5.611 -2.721 6.236 0.043 0.067
23 -1.764 -1.280 2.180 0.006 0.012
25 -1.325 1.602 2.079 0.008 0.014

25 0.000 0.000 0.000
24 1.333 -1.588 2.073 0.008 0.014
26 3.545 2.366 4.262 0.045 0.066
27 -4.877 -0.778 4.939 0.026 0.049

26 -3.500 -2.300 4.188
25 -3.500 -2.300 4.188 0.045 0.066

27 0.000 0.000 0.000
25 4.903 0.827 4.972 0.026 0.049
28 -18.184 -4.157 18.653 -0.000 1.309
29 6.189 1.668 6.410 0.086 0.162
30 7.091 1.661 7.283 0.161 0.304

28 0.000 0.000 0.000
27 18.184 5.466 18.987 -0.000 1.309 0.968
8 0.575 -1.999 2.080 0.000 -4.368
6 -18.759 -3.467 19.077 0.060 -13.086

29 -2.400 -0.900 2.563
27 -6.104 -1.506 6.286 0.086 0.162
30 3.704 0.606 3.753 0.033 0.063

30 -10.600 -1.900 10.769
27 -6.930 -1.358 7.062 0.161 0.304
29 -3.670 -0.542 3.710 0.033 0.063
Total loss 17.599 22.244
*** End of Experiment 5***


"Chapter: Balanced/ Unbalanced Fault Analysis "


A.1. A 3-phase transmission line operating at 33 KV & having a resistance
& reactance of 5 ohm & 20 ohm respectively. Is connected to a
generating station bus bar through a 15 MVA step up transformer which
has a reactance of 0.06 pu. Connected to the bus bar are two
generators, one 10 MVA having 0.1 pu reactance & other 5 MVA having
0.075 pu reactance. Calculate the short circuit MVA & the fault current
when a 3-phase short circuit occurs (a) at the HV terminals of
transformer (b) at the load end of the transmission line.
[Ref. 16.1, A.H]

A.2. A 33 KV bus bar has a 3-phase fault level of 1000 MVA. The negative &
zero sequence source reactance are 2/3 & 1/3 of positive sequence
reactance's. The zero sequence source resistance is 60 ohm. A 30 MVA,
33/132 KW solidly grounded delta/star transformer having a reactance of
0.1 pu is fed from 33 KV bus. Find the fault current & fault MVA at 132
KV bus for following faults (a) 3-phase (b) SLG (c) LL (d) DLG.
[Ref. 10.28 BRG]


B.1. For the shown system, a three-phase fault with fault impedance
Zf=j0.16 pu occurs at bus 3, simulate the problem in the computer & analyse
the result. [Ref. 9.6, H.S.]

Computer Program
>> zdata = [ 0 1 0 0.2
0 2 0 0.4
1 2 0 0.8
1 3 0 0.4
2 3 0 0.4];
OUTPUT
Zbus = zbuild(zdata)
symfault(zdata, Zbus)

Zbus =

0 + 0.1600i 0 + 0.0800i 0 + 0.1200i
0 + 0.0800i 0 + 0.2400i 0 + 0.1600i
0 + 0.1200i 0 + 0.1600i 0 + 0.3400i

Enter Faulted Bus No. -> 3
Enter Fault Impedance Zf = R + j*X in complex form (for bolted fault enter
0). Zf = 0+j*0.16
Balanced three-phase fault at bus No. 3
Total fault current = 2.0000 per unit
Bus Voltages during fault in per unit
Bus Voltage Angle
No. Magnitude degrees
1 0.7600 0.0000
2 0.6800 0.0000
3 0.3200 0.0000
Line currents for fault at bus No. 3
From To Current Angle
Bus Bus Magnitude degrees
G 1 1.2000 -90.0000
1 2 0.1000 -90.0000
1 3 1.1000 -90.0000
G 2 0.8000 -90.0000
2 3 0.9000 -90.0000
3 F 2.0000 -90.0000
Another fault location? Enter 'y' or 'n' within single quote -> 'n'

B.1. The one line diagram of a simple power system is shown. The neutral of
each generator is grounded through a current limiting reactor of 0.25/3 pu
on a 100 MVA base. The system data is given below. The generators are
running on no load at their rated voltage & rated frequency with their emfs
in phase. Determine the fault currents for the following faults
a) A balanced 3-phase fault at bus 3 through a fault impedance Zf= j0.1 pu.
b) A SLG fault at bus 3 through a fault impedance Zf=j0.1 pu.
c) A line to line fault at bus 3 through fault impedance Zf=j0.1 pu.
d) A LLG fault at bus 3 through a fault impedance Zf= j0.1 pu.

"Item "Base MVA "Voltage KV "X1 "X2 "X0 "
"G11 "100 "20 "0.15 "0.15 "0.05 "
"G2 "100 "20 "0.15 "0.15 "0.05 "
"T1 "100 "20/220 "0.1 "0.1 "0.10 "
"T2 "100 "20/220 "0.1 "0.1 "0.10 "
"L12 "100 "220 "0.125 "0.125 "0.3 "
"L13 "100 "220 "0.15 "0.15 "0.35 "
"L23 "100 "220 "0.25 "0.25 "0.7125 "

[Ref 10.7; H.S.]

Computer Program

>> zdata1 = [0 1 0 0.25
0 2 0 0.25
1 2 0 0.125
1 3 0 0.15
2 3 0 0.25];

zdata0 = [0 1 0 0.40
0 2 0 0.10
1 2 0 0.30
1 3 0 0.35
2 3 0 0.7125];
zdata2 = zdata1;
Zbus1 = zbuild(zdata1)
Zbus0 = zbuild(zdata0)
Zbus2 = Zbus1;
disp('(a) Symmetrical three-phase fault')
symfault(zdata1, Zbus1)
disp('(b) Line-to-ground fault' )
lgfault(zdata0, Zbus0, zdata1, Zbus1, zdata2, Zbus2)
disp('(c) Line-to-line fault')
llfault(zdata1, Zbus1, zdata2, Zbus2)
disp('(d) double line-to-ground fault')
dlgfault(zdata0, Zbus0, zdata1, Zbus1, zdata2, Zbus2)
OUTPUT
Zbus1 =

0 + 0.1450i 0 + 0.1050i 0 + 0.1300i
0 + 0.1050i 0 + 0.1450i 0 + 0.1200i
0 + 0.1300i 0 + 0.1200i 0 + 0.2200i

Zbus0 =

0 + 0.1820i 0 + 0.0545i 0 + 0.1400i
0 + 0.0545i 0 + 0.0864i 0 + 0.0650i
0 + 0.1400i 0 + 0.0650i 0 + 0.3500i

(a) Symmetrical three-phase fault

0). Zf = j*0.1

Balanced three-phase fault at bus No. 3

Bus Voltages during fault in per unit
Bus Voltage Angle
No. Magnitude degrees
1 0.5938 0.0000
2 0.6250 0.0000
3 0.3125 0.0000

From To Current Angle
Bus Bus Magnitude degrees
G 1 1.6250 -90.0000
1 3 1.8750 -90.0000
G 2 1.5000 -90.0000
2 1 0.2500 -90.0000
2 3 1.2500 -90.0000
3 F 3.1250 -90.0000
(b) Line-to-ground fault

Line-to-ground fault analysis

0). Zf = j*0.1

Single line to-ground fault at bus No. 3

Bus Voltages during the fault in per unit

Bus -------Voltage Magnitude-------
No. Phase a Phase b Phase c
1 0.6330 1.0046 1.0046
2 0.7202 0.9757 0.9757
3 0.2752 1.0647 1.0647


From To -----Line Current Magnitude----
Bus Bus Phase a Phase b Phase c
1 3 1.6514 0.0000 0.0000
2 1 0.3761 0.1560 0.1560
2 3 1.1009 0.0000 0.0000
3 F 2.7523 0.0000 0.0000
(c) Line-to-line fault

Line-to-line fault analysis

0). Zf = j*0.1

Line-to-line fault at bus No. 3


1 1.0000 0.6720 0.6720
2 1.0000 0.6939 0.6939
3 1.0000 0.5251 0.5251


1 3 0.0000 1.9245 1.9245
2 1 0.0000 0.2566 0.2566
2 3 0.0000 1.2830 1.2830
3 F 0.0000 3.2075 3.2075
(d) double line-to-ground fault

Double line-to-ground fault analysis

0). Zf = j*0.1

Double line-to-ground fault at bus No. 3


1 1.0066 0.5088 0.5088
2 0.9638 0.5740 0.5740
3 1.0855 0.1974 0.1974


1 3 0.0000 2.4350 2.4350
2 1 0.1118 0.3682 0.3682
2 3 0.0000 1.6233 1.6233
3 F 0.0000 4.0583 4.0583



Assignment/ Experiment No7

"Chapter: Equal Area Criterion "


A.1. For the system shown below, the pu values of different quantities are
E=1.2, V=1.0, Xd=0.2, X1=X2=0.4 the system is operating in equilibrium
with p1=1.5 pu. when one of the lines is suddenly switched ON. Predict
whether the system will be stable or not. If system is stable find the
maximum value which 'δ' attains. [Ex.13.5; BRG]

A.2. A 100 MVA, 50 HZ alternator is operating at Rated speed. The H
constant of the machine is KW/KVA. The load suddenly increases by 50
MW. Due to delay in governor action there is delay of 0.6 sec. is
operating of steam valve. Find the frequency deviation.

[Ex. 13.10, B.R.G]

A.3. A 3-phase generator derives 1.0 pu power to an infinite bus through a
transmission network when a fault occurs, the maximum power which
can be transferred prefault; during fault and post fault conditions is
1.7, 0.4, & 1.25 pu. Find critical clearing angle.[Ex.13.11;BRG]

A.4 A generating is delivering 0.6 of maximum power to an infinite bus
through transmission line. A fault occurs such that the reactance
between the generator & the infinite bus is increased to three times its
prefault value. When the fault is cleared, the maximum power that can be
delivered is 0.8 of the original maximum value. Determine the critical
clearing angle.

[Ref. 19.9, A.H.]


B.1. A 60 Hz synchronous generator having inertia constant H=9.94 MJ/MVA &
a transient reactance X'd=0.3 pu is connected to an infinite bus
through a purely reactive circuit as shown in fig. reactance's are
marked on the diag. on a common system base. The machine delivering a
real power of 0.6 pu, at 0.8 power factor lagging to the infinite bus
bar. The infinite bus bar voltage is 1.0 pu, determine (a) the maximum
power input that can be applied without loss of synchronism. (b)
Repeat (a) with zero initial power input. Assume the generator internal
voltage remains constant at the value computed in (a). The transfer
reactance & the generator internal voltage to be X=0.65 pu, & E'=1.35
pu. [Ex. 11.4;HS]

Computer Program

>> % (a) Initial real power P0 = 0.60
P0 = 0.6; E = 1.35; V = 1.0; X = 0.65;
eacpower(P0, E, V, X)
h=figure;
% (b) Zero initial power
P0 = 0;
eacpower(P0, E, V, X)

Initial power = 0.600 p.u.
Initial power angle = 16.791 degrees
Sudden additional power = 1.084 p.u.
Total power for critical stability = 1.684 p.u.
Maximum angle swing =125.840 degrees
New operating angle = 54.160 degrees

Current plot held

Initial power = 0.000 p.u.
Initial power angle = 0.000 degrees
Sudden additional power = 1.505 p.u.
Total power for critical stability = 1.505 p.u.
Maximum angle swing =133.563 degrees
New operating angle = 46.437 degrees

Current plot held

B.2. A 60 Hz synchronous generator having inertia constant H=5 MJ/MVA
& a direct axis transient reactance X'd=0.3 pu is connected to an
infinite bus through a purely reactive circuit as shown in fig.
reactance's are marked on the diag. on common system base. The
generator is delivering real power Pe=0.8 pu & Q=0.074 pu to infinite
bus at a voltage of V=1 pu (a) A temporary 3-phase fault occurs at the
sending end of the line at point F. when the fault is cleared, both
lines are intact. Determine the critical clearing angle & the critical
fault clearing time. (b) A 3-phase fault occurs at the sending of the
lines, the fault is cleared, & the faulted line is isolated. Determine
the critical clearing angle. [Ex. 11.5;HS].

Computer Program

>> %(a) Fault at the sending end. Both lines intact when fault is cleared
Pm = 0.8; E= 1.17; V = 1.0; X1 = 0.65; X2 = inf; X3 = 0.65;
eacfault(Pm, E, V, X1, X2, X3)
OUTPUT
%(b) Fault at the mid-point of one line. Faulted line is isolated
X2 = 1.8; X3 = 0.8;
eacfault(Pm, E, V, X1, X2, X3)

For this case tc can be found from analytical formula.
To find tc enter Inertia Constant H, (or 0 to skip) H = 5

Initial power angle = 26.388
Maximum angle swing = 153.612
Critical clearing angle = 84.775

Critical clearing time = 0.260 sec.

Current plot held

Initial power angle = 26.388
Maximum angle swing = 146.838
Critical clearing angle = 98.834
Current plot held




"Chapter:Transiant Stability Analysis "


A.1. A 50 MVA, 50 HZ generator delivers 50 MW over a double circuit line
to an infinite bus. The generator has an inertia constant H=2.7 MJ/MVA.
The generator has an reactance Xd=0.3 pu. Each transmission line has a
reactance of 0.5 pu on same base. E=1.4 pu. & V=1.0 00 pu. A solid 3-
phase fault to ground occurs on line. The system reactances during the
fault are shown in figure.

Plot the Swing curve for a sustained fault. [ Ex 19.11; A.H.]


B.1.A 60 Hz synchronous generator having inertia constant H=5 MJ/MVA & a
direct axis transient reactance X'd=0.3 pu is connected to an infinite
bus through a purely reactive circuit as shown in fig. reactance's are
marked on the diag. on common system base. The generator is delivering
real power Pe=0.8 pu & Q=0.074 pu to infinite bus at a voltage of V=1
pu.A 3-phase fault at the middle of one line is cleared by isolating the
faulted circuit simultaneously at both ends.(a) The fault is cleared in
30 second. Obtain the numerical solution of the swing equation for
1.0sec. using the modified Euler method with a step size of t=0.01 sec.
From the swing curve determine the system stability. (b) Repeat the
simulation & obtain the swing plots for the critical clearing time, &
when fault is cleared in 0.5 sec. [ Ex.11.6; H.S.]
Computer Program

global Pm f H E V X1 X2 X3
Pm = 0.80; E = 1.17; V = 1.0;
X1 = 0.65; X2 = 1.80; X3 = 0.8;
H = 5.0; f = 60; tf = 1; Dt = 0.01;
% (a) Fault is cleared in 0.3 sec.
tc = 0.3;
swingmeu(Pm, E, V, X1, X2, X3, H, f, tc, tf, Dt)
% (b) Fault is cleared in 0.4 sec. and 0.5 sec.
tc = .5;
tc = .4;

disp('Parts (a) & (b) are repeated using swingrk4')
disp('Press Enter to continue')
pause
tc = 0.3;
swingrk4(Pm, E, V, X1, X2, X3, H, f, tc, tf)

tc = .5;
tc = .4;
Output
"Fault is cleared at " 0.4200 101.0221 " 0.8500 122.1239 "
"0.400 Sec. "4.4322 "-2.0759 "
"time delta Dw "0.4300 103.4929 "0.8600 120.8871 "
" "4.1952 "-2.2442 "
"s degrees rad/s"0.4400 105.8294 "0.8700 119.5521 "
"0 26.3878 0 "3.9636 "-2.4189 "
"0.0100 26.4430 "0.4500 108.0349 "0.8800 118.1152 "
"0.1927 "3.7380 "-2.6001 "
"0.0200 26.6085 "0.4600 110.1128 "0.8900 116.5725 "
"0.3849 "3.5187 "-2.7881 "
"0.0300 26.8841 "0.4700 112.0669 "0.9000 114.9202 "
"0.5764 "3.3060 "-2.9830 "
"0.0400 27.2689 "0.4800 113.9011 "0.9100 113.1542 "
"0.7665 "3.1002 "-3.1848 "
"0.0500 27.7624 "0.4900 115.6194 "0.9200 111.2706 "
"0.9550 "2.9013 "-3.3934 "
"0.0600 28.3632 "0.5000 117.2258 "0.9300 109.2655 "
"1.1414 "2.7094 "-3.6089 "
"0.0700 29.0703 "0.5100 118.7241 "0.9400 107.1351 "
"1.3254 "2.5242 "-3.8309 "
"0.0800 29.8820 "0.5200 120.1182 "0.9500 104.8756 "
"1.5065 "2.3457 "-4.0591 "
"0.0900 30.7966 "0.5300 121.4120 "0.9600 102.4837 "
"1.6844 "2.1737 "-4.2930 "
"0.1000 31.8121 "0.5400 122.6090 "0.9700 99.9561 "
"1.8588 "2.0079 "-4.5320 "
"0.1100 32.9265 "0.5500 123.7128 "0.9800 97.2903 "
"2.0293 "1.8481 "-4.7754 "
"0.1200 34.1374 "0.5600 124.7267 "0.9900 94.4839 "
"2.1956 "1.6939 "-5.0220 "
"0.1300 35.4424 "0.5700 125.6538 "1.0000 91.5354 "
"2.3575 "1.5451 "-5.2708 "
"0.1400 36.8389 "0.5800 126.4971 " "
"2.5146 "1.4012 " "
"0.1500 38.3240 "0.5900 127.2594 " "
"2.6669 "1.2619 " "
"0.1600 39.8948 "0.6000 127.9430 " "
"2.8140 "1.1268 " "
"0.1700 41.5485 "0.6100 128.5505 " "
"2.9558 "0.9955 " "
"0.1800 43.2819 "0.6200 129.0837 " "
"3.0922 "0.8676 " "
"0.1900 45.0918 "0.6300 129.5446 " "
"3.2230 "0.7427 " "
"0.2000 46.9752 "0.6400 129.9347 " "
"3.3483 "0.6205 " "
"0.2100 48.9287 "0.6500 130.2555 " "
"3.4681 "0.5004 " "
"0.2200 50.9492 "0.6600 130.5081 " "
"3.5822 "0.3821 " "
"0.2300 53.0336 "0.6700 130.6933 " "
"3.6908 "0.2652 " "
"0.2400 55.1785 "0.6800 130.8119 " "
"3.7939 "0.1492 " "
"0.2500 57.3810 "0.6900 130.8643 " "
"3.8918 "0.0338 " "
"0.2600 59.6381 "0.7000 130.8506 " "
"3.9845 "-0.0815 " "
"0.2700 61.9469 "0.7100 130.7708 " "
"4.0722 "-0.1971 " "
"0.2800 64.3046 "0.7200 130.6246 " "
"4.1553 "-0.3134 " "
"0.2900 66.7085 "0.7300 130.4116 " "
"4.2340 "-0.4309 " "
"0.3000 69.1564 "0.7400 130.1308 " "
"4.3086 "-0.5499 " "
"0.3100 71.6458 "0.7500 129.7814 " "
"4.3794 "-0.6708 " "
"0.3200 74.1747 "0.7600 129.3621 " "
"4.4468 "-0.7941 " "
"0.3300 76.7414 "0.7700 128.8714 " "
"4.5113 "-0.9201 " "
"0.3400 79.3443 "0.7800 128.3077 " "
"4.5732 "-1.0494 " "
"0.3500 81.9820 "0.7900 127.6689 " "
"4.6331 "-1.1822 " "
"0.3600 84.6534 "0.8000 126.9529 " "
"4.6914 "-1.3190 " "
"0.3700 87.3579 "0.8100 126.1573 " "
"4.7486 "-1.4602 " "
"0.3800 90.0949 "0.8200 125.2796 " "
"4.8053 "-1.6061 " "
"0.3900 92.8643 "0.8300 124.3168 " "
"4.8620 "-1.7571 " "
"0.4000 95.6663 "0.8400 123.2660 " "
"4.9193 "-1.9136 " "
"0.4100 98.4140 " " "
"4.6739 " " "

Searching with a final time of 0.80 Sec.

Critical clearing time = 0.41 seconds
Critical clearing angle = 98.83 degrees

Parts (a) & (b) are repeated using swingrk4
Press Enter to continue
"Fault is cleared at " 0.1407 36.9382 " 0.5600 124.6751 "
"0.400 Sec. "2.5252 "1.6884 "
" "0.1507 38.4288 "0.5750 126.0300 "
"time delta Dw "2.6770 "1.4664 "
" "0.1607 40.0049 "0.5900 127.1987 "
"s degrees rad/s"2.8237 "1.2551 "
" "0.1707 41.6635 "0.6050 128.1900 "
"0 26.3878 0 "2.9651 "1.0532 "
"0.0000 26.3878 "0.1807 43.4016 "0.6200 129.0113 "
"0.0001 "3.1011 "0.8594 "
"0.0000 26.3878 "0.1907 45.2160 "0.6350 129.6691 "
"0.0001 "3.2315 "0.6723 "
"0.0000 26.3878 "0.2007 47.1036 "0.6500 130.1685 "
"0.0002 "3.3564 "0.4905 "
"0.0000 26.3878 "0.2107 49.0611 "0.6650 130.5133 "
"0.0002 "3.4757 "0.3127 "
"0.0000 26.3878 "0.2207 51.0853 "0.6800 130.7065 "
"0.0005 "3.5894 "0.1373 "
"0.0000 26.3878 "0.2307 53.1731 "0.6950 130.7497 "
"0.0007 "3.6976 "-0.0369 "
"0.0000 26.3878 "0.2407 55.3214 "0.7100 130.6431 "
"0.0010 "3.8003 "-0.2114 "
"0.0001 26.3878 "0.2507 57.5270 "0.7250 130.3858 "
"0.0012 "3.8978 "-0.3877 "
"0.0001 26.3878 "0.2607 59.7869 "0.7400 129.9758 "
"0.0025 "3.9901 "-0.5670 "
"0.0002 26.3878 "0.2707 62.0983 "0.7550 129.4099 "
"0.0037 "4.0775 "-0.7509 "
"0.0003 26.3878 "0.2807 64.4585 "0.7700 128.6835 "
"0.0050 "4.1602 "-0.9406 "
"0.0003 26.3879 "0.2907 66.8648 "0.7850 127.7909 "
"0.0062 "4.2386 "-1.1377 "
"0.0006 26.3880 "0.3007 69.3148 "0.8000 126.7254 "
"0.0125 "4.3129 "-1.3434 "
"0.0010 26.3883 "0.3107 71.8063 "0.8150 125.4789 "
"0.0188 "4.3834 "-1.5591 "
"0.0013 26.3887 "0.3207 74.3372 "0.8300 124.0423 "
"0.0251 "4.4506 "-1.7860 "
"0.0016 26.3893 "0.3307 76.9057 "0.8450 122.4054 "
"0.0313 "4.5148 "-2.0253 "
"0.0033 26.3937 "0.3407 79.5103 "0.8600 120.5572 "
"0.0627 "4.5766 "-2.2780 "
"0.0049 26.4010 "0.3507 82.1497 "0.8750 118.4857 "
"0.0941 "4.6363 "-2.5451 "
"0.0065 26.4112 "0.3607 84.8228 "0.8900 116.1782 "
"0.1255 "4.6944 "-2.8272 "
"0.0081 26.4244 "0.3706 87.4789 "0.9050 113.6216 "
"0.1569 "4.7505 "-3.1248 "
"0.0163 26.5343 "0.3804 90.1663 "0.9200 110.8026 "
"0.3136 "4.8061 "-3.4379 "
"0.0244 26.7172 "0.3902 92.8850 "0.9350 107.7081 "
"0.4699 "4.8617 "-3.7660 "
"0.0326 26.9729 "0.4000 95.6351 "0.9500 104.3252 "
"0.6255 "4.9179 "-4.1083 "
"0.0407 27.3010 "0.4000 95.6351 "0.9650 100.6427 "
"0.7802 "4.9179 "-4.4632 "
"0.0507 27.8021 "0.4100 98.3820 "0.9800 96.6506 "
"0.9685 "4.6724 "-4.8282 "
"0.0607 28.4104 "0.4200 100.9893 "0.9850 95.2495 "
"1.1546 "4.4306 "-4.9516 "
"0.0707 29.1247 "0.4300 103.4594 "0.9900 93.8128 "
"1.3383 "4.1934 "-5.0757 "
"0.0807 29.9435 "0.4400 105.7951 "0.9950 92.3406 "
"1.5192 "3.9616 "-5.2002 "
"0.0907 30.8649 "0.4550 109.0542 "1.0000 90.8327 "
"1.6968 "3.6251 "-5.3250 "
"0.1007 31.8871 "0.4700 112.0305 "Searching with a final "
"1.8708 "3.3033 "time of 0.80 Sec. "
"0.1107 33.0080 "0.4850 114.7366 "Critical clearing time ="
"2.0410 "2.9968 "0.41 seconds "
"0.1207 34.2251 "0.5000 117.1860 "Critical clearing angle "
"2.2069 "2.7058 "= 98.83 degrees "
"0.1307 35.5361 "0.5150 119.3919 " "
"2.3684 "2.4301 " "
" "0.5300 121.3673 " "
" "2.1692 " "
" "0.5450 123.1245 " "
" "1.9223 " "



Experiment No. 9

"Chapter: Load Frequency Control "

A. NIL

B.1 An isolated power station has the following parameters Turbine
time constant Tt=0.5 sec.; Governor time constant Tg =0.2 sec.
Generator time constant H =5 sec; Governor speed regulation = R pu; The
load varies by 0.8% for 1 % for a 1% change in frequency, i.e. D=0.8
(a) use the Routh-Hurwith array to find the range of R for control
system stability. (b) Use MATLB rlocus function to obtain the rootlocus
plot. (c) The governor speed regulation is set to R=0.05 pu. The
turbine rated output is 250 MW at nominal frequency of 60 Hz. A sudden
load change of 50 MW 1( PL=0.2 pu) occurs.[i] find the steady-state
frequency deviation in Hz.[ii] use MATLB to obtain the time-domain
performance specifications & the frequency deviation step response.[Ex.
12.1; H.S.]

Computer Program
>> disp('Example 12.1 (b) Root-locus')
num = 1;
den = [1 7.08 10.56 .8];
figure (1), rlocus(num, den)
disp('Example 12.1 (c) Frequency deviation step response')
PL = 0.2;
numc = [0.1 0.7 1];
denc = [1 7.08 10.56 20.8];
t = 0:.02:10;
c = -PL*step(numc, denc, t);
figure(2), plot(t, c), grid
xlabel('t, sec'), ylabel('pu')
title('Frequency deviation step response')
timespec(numc, denc)
Output
Example 12.1 (b) Root-locus
Example 12.1 (c) Frequency deviation step response
Peak time = 1.22311 Percent overshoot = 54.8019
Rise time = 0.418873
Settling time = 6.80249

A.2An isolated power station has the following parameters
Turbine time constant Tt=0.5 sec.; Governor time constant Tg =0.2
sec.
Generator time constant H =5 sec; Governor speed regulation = R
pu; The load varies by 0.8% for 1 % for a 1% change in
frequency, i.e. D=0.8. this is equipped with the secondary integral
control loop for automatic generation control. (a) Use the MATLAB to
obtain the frequency deviation step response for a sudden load change of
PL=0.2 pu. Set the integral controller gain to KI = 7.
[Ex. 12.6; H.S]

Computer Program
>> PL = 0.2;
KI = 7;
numc = [0.1 0.7 1 0];
denc = [1 7.08 10.56 20.8 KI];
t = 0:.02:12;
c=-PL*step(numc, denc, t);
plot(t, c), grid
xlabel('t, sec'), ylabel('pu')
title('Frequency deviation step response')

OUTPUT

A.3. The AVR system of a generator has following parameters

" "Gain "Time Constant "
"Amplifier "KA "TA = 0.1 "
"Exciter "KE =1 "TE = 0.4 "
"Generator "KG =1 "TG = 1 "
"Sensor "KR =1 "TR = 0.05 "

a) Use the Routh-Hurwitz array to find the range of KA for control system
stability.
b) Use MATLAB to obtain the rootlocus plot.
c) The amplifier gain is set to KA = 10;
i. Find the steady-state step response.
ii. Use MATLAB to obtain the step response & the time domain
performance specifications.

Computer Program
>> num=500;
den=[1 33.5 307.5 775 500];
figure(1), rlocus(num, den);

kA= 10;
numc=kA*[25 500];
denc=[1 33.5 307.5 775 500+ 500*kA];
t=0:.05:20;
c=step(numc, denc, t);
figure(2), plot(t, c), xlabel('t, sec'), grid
timespec(numc, denc)
disp('(d) Open sim12ex6.mdl in SIMULINK WINDOW and click on simulation')

OUTPUT
Peak time = 0.791532 Percent overshoot = 82.4694
Rise time = 0.247354
Settling time = 19.0462

d) Open sim12ex6.mdl in SIMULINK WINDOW and click on simulation


Steps to Installing MATLAB programs

"Step 1 "Create folder 'power' in MATLAB6p5 toolbox "
" "i.e. 'C\Matlb6p5\toolbox\power' "
"Step 2 "Copy 'Folder 1' & 'Folder 2' from 'power system toolbox' to "
" "'Power' folder, which is created in step 1 "
"Step 3 "Open MATLAB command window. "
"Step 4 "Click 'set path' from 'File' menu "
"Step 5 "From path menu window click on 'Add folder' button. This will "
" "browse for folder window opens. "
"Step 6 "Select 'C\matlab6p5\toolbox\power' then click on OK. 'Add' "
" "this folder. You will see this folder is added. "
"Step 7 "Press 'save' button & Press 'close'. "

Power system toolbox diskette is available with 'Power System Analysis, by
Hadi Sadat" book.

How to use: The file name for chapter examples begins with the letter
'chp'.

For example program file for Example 10.4 is 'ch10ex4'.

-----------------------
Manual for

POWER SYSTEM

ANALYSIS & CONTROL

j[?]j[?]7j[?]9j[?]:j[?];j[?] m[?]0m[?]`m[?]¦m[?]Âm[?]øm[?]

n[?],n[?]Wn[?]úúòúúúúòúúúéàúÓúúúúúúúúú

&dPÆÿgdã>È Â^ ÂT.E. PART-II (6TH Semester)

Electrical Engineering

Shivaji University, Kolhapur

Load

1

2

-J 3

-j5

-j4

3

Load

1.5+JQG22

1+J0.5

Pv bUS

Slack bus

PQ -bus

1.2+J0.5

Psac manual

Recommend Documents